Question: Divide the following rational expressions and simplify the result. $\dfrac{6y^2+13y+6}{20-5y} \div \dfrac{4y^2-9}{y^2-2y-8}=$
Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $6y^2+13y+6$, of the first expression can be factored by grouping to $(3y+2)(2y+3)$. The denominator, $20-5y$, of the first expression can be factored as $-5(y-4)$ by factoring out $-5$. The numerator, $4y^2-9$, of the second expression can be factored as $(2y+3)(2y-3)$ using the difference of squares pattern. The denominator, $y^2-2y-8$, of the second expression can be factored as $(y-4)(y+2)$ using the sum-product pattern. Now the quotient looks as follows: $\dfrac{(3y+2)(2y+3)}{-5(y-4)} \div \dfrac{(2y+3)(2y-3)}{(y-4)(y+2)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{(3y+2)(2y+3)}{-5(y-4)} \div \dfrac{(2y+3)(2y-3)}{(y-4)(y+2)}$ $\begin{aligned} &= \dfrac{(3y+2)(2y+3)}{-5(y-4)} \cdot \dfrac{(y-4)(y+2)}{(2y+3)(2y-3)} &\text{Flip the divisor.}\\\\\\ &= \dfrac{(3y+2)(2y+3)\cdot(y-4)(y+2)}{-5(y-4)\cdot(2y+3)(2y-3)} &\text{Multiply across.}\\\\\\ &= \dfrac{(3y+2){\cancel{(2y+3)}}{\cancel{(y-4)}}(y+2)}{-5 {\cancel{(y-4)}} {\cancel{(2y+3)}}(2y-3)} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(3y+2)(y+2)}{-5(2y-3)} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{(3y+2)(y+2)}{-5(2y-3)}$, which is equivalent to $\dfrac{3y^2+8y+4}{15-10y}$.